3.8.67 \(\int \frac {1}{x^5 (a+b x^2+c x^4)^{3/2}} \, dx\)

Optimal. Leaf size=195 \[ -\frac {3 \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{16 a^{7/2}}+\frac {b \left (15 b^2-52 a c\right ) \sqrt {a+b x^2+c x^4}}{8 a^3 x^2 \left (b^2-4 a c\right )}-\frac {\left (5 b^2-12 a c\right ) \sqrt {a+b x^2+c x^4}}{4 a^2 x^4 \left (b^2-4 a c\right )}+\frac {-2 a c+b^2+b c x^2}{a x^4 \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}} \]

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Rubi [A]  time = 0.21, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {1114, 740, 834, 806, 724, 206} \begin {gather*} \frac {b \left (15 b^2-52 a c\right ) \sqrt {a+b x^2+c x^4}}{8 a^3 x^2 \left (b^2-4 a c\right )}-\frac {\left (5 b^2-12 a c\right ) \sqrt {a+b x^2+c x^4}}{4 a^2 x^4 \left (b^2-4 a c\right )}-\frac {3 \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{16 a^{7/2}}+\frac {-2 a c+b^2+b c x^2}{a x^4 \left (b^2-4 a c\right ) \sqrt {a+b x^2+c x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^2 + c*x^4)^(3/2)),x]

[Out]

(b^2 - 2*a*c + b*c*x^2)/(a*(b^2 - 4*a*c)*x^4*Sqrt[a + b*x^2 + c*x^4]) - ((5*b^2 - 12*a*c)*Sqrt[a + b*x^2 + c*x
^4])/(4*a^2*(b^2 - 4*a*c)*x^4) + (b*(15*b^2 - 52*a*c)*Sqrt[a + b*x^2 + c*x^4])/(8*a^3*(b^2 - 4*a*c)*x^2) - (3*
(5*b^2 - 4*a*c)*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(16*a^(7/2))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 740

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
b*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e
+ a*e^2)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b
*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 834

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p])

Rule 1114

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*(a +
b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {1}{x^5 \left (a+b x^2+c x^4\right )^{3/2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{x^3 \left (a+b x+c x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=\frac {b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^4 \sqrt {a+b x^2+c x^4}}-\frac {\operatorname {Subst}\left (\int \frac {\frac {1}{2} \left (-5 b^2+12 a c\right )-2 b c x}{x^3 \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{a \left (b^2-4 a c\right )}\\ &=\frac {b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^4 \sqrt {a+b x^2+c x^4}}-\frac {\left (5 b^2-12 a c\right ) \sqrt {a+b x^2+c x^4}}{4 a^2 \left (b^2-4 a c\right ) x^4}+\frac {\operatorname {Subst}\left (\int \frac {-\frac {1}{4} b \left (15 b^2-52 a c\right )-\frac {1}{2} c \left (5 b^2-12 a c\right ) x}{x^2 \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{2 a^2 \left (b^2-4 a c\right )}\\ &=\frac {b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^4 \sqrt {a+b x^2+c x^4}}-\frac {\left (5 b^2-12 a c\right ) \sqrt {a+b x^2+c x^4}}{4 a^2 \left (b^2-4 a c\right ) x^4}+\frac {b \left (15 b^2-52 a c\right ) \sqrt {a+b x^2+c x^4}}{8 a^3 \left (b^2-4 a c\right ) x^2}+\frac {\left (3 \left (5 b^2-4 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx,x,x^2\right )}{16 a^3}\\ &=\frac {b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^4 \sqrt {a+b x^2+c x^4}}-\frac {\left (5 b^2-12 a c\right ) \sqrt {a+b x^2+c x^4}}{4 a^2 \left (b^2-4 a c\right ) x^4}+\frac {b \left (15 b^2-52 a c\right ) \sqrt {a+b x^2+c x^4}}{8 a^3 \left (b^2-4 a c\right ) x^2}-\frac {\left (3 \left (5 b^2-4 a c\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 a-x^2} \, dx,x,\frac {2 a+b x^2}{\sqrt {a+b x^2+c x^4}}\right )}{8 a^3}\\ &=\frac {b^2-2 a c+b c x^2}{a \left (b^2-4 a c\right ) x^4 \sqrt {a+b x^2+c x^4}}-\frac {\left (5 b^2-12 a c\right ) \sqrt {a+b x^2+c x^4}}{4 a^2 \left (b^2-4 a c\right ) x^4}+\frac {b \left (15 b^2-52 a c\right ) \sqrt {a+b x^2+c x^4}}{8 a^3 \left (b^2-4 a c\right ) x^2}-\frac {3 \left (5 b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{16 a^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 179, normalized size = 0.92 \begin {gather*} \frac {3 \left (16 a^2 c^2-24 a b^2 c+5 b^4\right ) \tanh ^{-1}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )+\frac {2 \sqrt {a} \left (-8 a^3 c+2 a^2 \left (b^2+10 b c x^2-12 c^2 x^4\right )+a b x^2 \left (-5 b^2+62 b c x^2+52 c^2 x^4\right )-15 b^3 x^4 \left (b+c x^2\right )\right )}{x^4 \sqrt {a+b x^2+c x^4}}}{16 a^{7/2} \left (4 a c-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a + b*x^2 + c*x^4)^(3/2)),x]

[Out]

((2*Sqrt[a]*(-8*a^3*c - 15*b^3*x^4*(b + c*x^2) + 2*a^2*(b^2 + 10*b*c*x^2 - 12*c^2*x^4) + a*b*x^2*(-5*b^2 + 62*
b*c*x^2 + 52*c^2*x^4)))/(x^4*Sqrt[a + b*x^2 + c*x^4]) + 3*(5*b^4 - 24*a*b^2*c + 16*a^2*c^2)*ArcTanh[(2*a + b*x
^2)/(2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(16*a^(7/2)*(-b^2 + 4*a*c))

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IntegrateAlgebraic [A]  time = 0.87, size = 175, normalized size = 0.90 \begin {gather*} \frac {-8 a^3 c+2 a^2 b^2+20 a^2 b c x^2-24 a^2 c^2 x^4-5 a b^3 x^2+62 a b^2 c x^4+52 a b c^2 x^6-15 b^4 x^4-15 b^3 c x^6}{8 a^3 x^4 \left (4 a c-b^2\right ) \sqrt {a+b x^2+c x^4}}-\frac {3 \left (4 a c-5 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x^2-\sqrt {a+b x^2+c x^4}}{\sqrt {a}}\right )}{8 a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(x^5*(a + b*x^2 + c*x^4)^(3/2)),x]

[Out]

(2*a^2*b^2 - 8*a^3*c - 5*a*b^3*x^2 + 20*a^2*b*c*x^2 - 15*b^4*x^4 + 62*a*b^2*c*x^4 - 24*a^2*c^2*x^4 - 15*b^3*c*
x^6 + 52*a*b*c^2*x^6)/(8*a^3*(-b^2 + 4*a*c)*x^4*Sqrt[a + b*x^2 + c*x^4]) - (3*(-5*b^2 + 4*a*c)*ArcTanh[(Sqrt[c
]*x^2 - Sqrt[a + b*x^2 + c*x^4])/Sqrt[a]])/(8*a^(7/2))

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fricas [A]  time = 2.74, size = 615, normalized size = 3.15 \begin {gather*} \left [-\frac {3 \, {\left ({\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{8} + {\left (5 \, b^{5} - 24 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{6} + {\left (5 \, a b^{4} - 24 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} x^{4}\right )} \sqrt {a} \log \left (-\frac {{\left (b^{2} + 4 \, a c\right )} x^{4} + 8 \, a b x^{2} + 4 \, \sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{4}}\right ) - 4 \, {\left ({\left (15 \, a b^{3} c - 52 \, a^{2} b c^{2}\right )} x^{6} - 2 \, a^{3} b^{2} + 8 \, a^{4} c + {\left (15 \, a b^{4} - 62 \, a^{2} b^{2} c + 24 \, a^{3} c^{2}\right )} x^{4} + 5 \, {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{32 \, {\left ({\left (a^{4} b^{2} c - 4 \, a^{5} c^{2}\right )} x^{8} + {\left (a^{4} b^{3} - 4 \, a^{5} b c\right )} x^{6} + {\left (a^{5} b^{2} - 4 \, a^{6} c\right )} x^{4}\right )}}, \frac {3 \, {\left ({\left (5 \, b^{4} c - 24 \, a b^{2} c^{2} + 16 \, a^{2} c^{3}\right )} x^{8} + {\left (5 \, b^{5} - 24 \, a b^{3} c + 16 \, a^{2} b c^{2}\right )} x^{6} + {\left (5 \, a b^{4} - 24 \, a^{2} b^{2} c + 16 \, a^{3} c^{2}\right )} x^{4}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {c x^{4} + b x^{2} + a} {\left (b x^{2} + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{4} + a b x^{2} + a^{2}\right )}}\right ) + 2 \, {\left ({\left (15 \, a b^{3} c - 52 \, a^{2} b c^{2}\right )} x^{6} - 2 \, a^{3} b^{2} + 8 \, a^{4} c + {\left (15 \, a b^{4} - 62 \, a^{2} b^{2} c + 24 \, a^{3} c^{2}\right )} x^{4} + 5 \, {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x^{2}\right )} \sqrt {c x^{4} + b x^{2} + a}}{16 \, {\left ({\left (a^{4} b^{2} c - 4 \, a^{5} c^{2}\right )} x^{8} + {\left (a^{4} b^{3} - 4 \, a^{5} b c\right )} x^{6} + {\left (a^{5} b^{2} - 4 \, a^{6} c\right )} x^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(c*x^4+b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/32*(3*((5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*x^8 + (5*b^5 - 24*a*b^3*c + 16*a^2*b*c^2)*x^6 + (5*a*b^4 - 24
*a^2*b^2*c + 16*a^3*c^2)*x^4)*sqrt(a)*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 + 4*sqrt(c*x^4 + b*x^2 + a)*(b*x^2 +
2*a)*sqrt(a) + 8*a^2)/x^4) - 4*((15*a*b^3*c - 52*a^2*b*c^2)*x^6 - 2*a^3*b^2 + 8*a^4*c + (15*a*b^4 - 62*a^2*b^
2*c + 24*a^3*c^2)*x^4 + 5*(a^2*b^3 - 4*a^3*b*c)*x^2)*sqrt(c*x^4 + b*x^2 + a))/((a^4*b^2*c - 4*a^5*c^2)*x^8 + (
a^4*b^3 - 4*a^5*b*c)*x^6 + (a^5*b^2 - 4*a^6*c)*x^4), 1/16*(3*((5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*x^8 + (5*b
^5 - 24*a*b^3*c + 16*a^2*b*c^2)*x^6 + (5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2)*x^4)*sqrt(-a)*arctan(1/2*sqrt(c*x^
4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(-a)/(a*c*x^4 + a*b*x^2 + a^2)) + 2*((15*a*b^3*c - 52*a^2*b*c^2)*x^6 - 2*a^3*
b^2 + 8*a^4*c + (15*a*b^4 - 62*a^2*b^2*c + 24*a^3*c^2)*x^4 + 5*(a^2*b^3 - 4*a^3*b*c)*x^2)*sqrt(c*x^4 + b*x^2 +
a))/((a^4*b^2*c - 4*a^5*c^2)*x^8 + (a^4*b^3 - 4*a^5*b*c)*x^6 + (a^5*b^2 - 4*a^6*c)*x^4)]

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giac [A]  time = 0.38, size = 350, normalized size = 1.79 \begin {gather*} \frac {\frac {{\left (a^{3} b^{3} c - 3 \, a^{4} b c^{2}\right )} x^{2}}{a^{6} b^{2} - 4 \, a^{7} c} + \frac {a^{3} b^{4} - 4 \, a^{4} b^{2} c + 2 \, a^{5} c^{2}}{a^{6} b^{2} - 4 \, a^{7} c}}{\sqrt {c x^{4} + b x^{2} + a}} + \frac {3 \, {\left (5 \, b^{2} - 4 \, a c\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{3}} - \frac {7 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} b^{2} - 4 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{3} a c + 8 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} a b \sqrt {c} - 9 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a b^{2} - 4 \, {\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} a^{2} c - 16 \, a^{2} b \sqrt {c}}{8 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )}^{2} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(c*x^4+b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

((a^3*b^3*c - 3*a^4*b*c^2)*x^2/(a^6*b^2 - 4*a^7*c) + (a^3*b^4 - 4*a^4*b^2*c + 2*a^5*c^2)/(a^6*b^2 - 4*a^7*c))/
sqrt(c*x^4 + b*x^2 + a) + 3/8*(5*b^2 - 4*a*c)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(
-a)*a^3) - 1/8*(7*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*b^2 - 4*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^3*
a*c + 8*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2*a*b*sqrt(c) - 9*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a*b^
2 - 4*(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*a^2*c - 16*a^2*b*sqrt(c))/(((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 +
a))^2 - a)^2*a^3)

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maple [A]  time = 0.02, size = 314, normalized size = 1.61 \begin {gather*} \frac {13 b \,c^{2} x^{2}}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{2}}-\frac {15 b^{3} c \,x^{2}}{8 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{3}}+\frac {13 b^{2} c}{4 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{2}}-\frac {15 b^{4}}{16 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{3}}+\frac {3 c \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{4 a^{\frac {5}{2}}}-\frac {15 b^{2} \ln \left (\frac {b \,x^{2}+2 a +2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}}{x^{2}}\right )}{16 a^{\frac {7}{2}}}-\frac {3 c}{4 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{2}}+\frac {15 b^{2}}{16 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{3}}+\frac {5 b}{8 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a^{2} x^{2}}-\frac {1}{4 \sqrt {c \,x^{4}+b \,x^{2}+a}\, a \,x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(c*x^4+b*x^2+a)^(3/2),x)

[Out]

-1/4/a/x^4/(c*x^4+b*x^2+a)^(1/2)+5/8*b/a^2/x^2/(c*x^4+b*x^2+a)^(1/2)+15/16*b^2/a^3/(c*x^4+b*x^2+a)^(1/2)-15/8*
b^3/a^3/(4*a*c-b^2)/(c*x^4+b*x^2+a)^(1/2)*x^2*c-15/16*b^4/a^3/(4*a*c-b^2)/(c*x^4+b*x^2+a)^(1/2)-15/16*b^2/a^(7
/2)*ln((b*x^2+2*a+2*(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)+13/2*b/a^2*c^2/(4*a*c-b^2)/(c*x^4+b*x^2+a)^(1/2)*x^2+1
3/4*b^2/a^2*c/(4*a*c-b^2)/(c*x^4+b*x^2+a)^(1/2)-3/4*c/a^2/(c*x^4+b*x^2+a)^(1/2)+3/4*c/a^(5/2)*ln((b*x^2+2*a+2*
(c*x^4+b*x^2+a)^(1/2)*a^(1/2))/x^2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(c*x^4+b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{x^5\,{\left (c\,x^4+b\,x^2+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^5*(a + b*x^2 + c*x^4)^(3/2)),x)

[Out]

int(1/(x^5*(a + b*x^2 + c*x^4)^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{5} \left (a + b x^{2} + c x^{4}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(c*x**4+b*x**2+a)**(3/2),x)

[Out]

Integral(1/(x**5*(a + b*x**2 + c*x**4)**(3/2)), x)

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